RF Cafe Software
About RF Cafe
Copyright
1996 
2022
Webmaster:
Kirt Blattenberger,
BSEE
 KB3UON
RF Cafe began life in 1996 as "RF Tools" in an AOL screen name web space totaling 2 MB. Its primary purpose was to provide me with ready access to commonly needed formulas and reference material while performing my work as an RF system and circuit design engineer. The Internet was still largely an unknown entity at the time and not much was available in the form of WYSIWYG ...
All trademarks, copyrights, patents, and other rights of ownership to images and text used on the RF Cafe website are hereby acknowledged.
My Hobby Website:
AirplanesAndRockets.com
Try Using SEARCH
to
Find What You Need.
There are 1,000s of Pages Indexed on RF Cafe !
Two Conductors in Parallel (Unbalanced) Above Ground Plane For D << d, h Z_{0}= (69/ε^{½}) log_{10}{(4h/d)[1+(2h/D)^{2}]^{½}} 
Single Conductor Above Ground Plane For d << h Z_{0}= (138/ε^{½}) log_{10}(4h/d) 
Two Conductors in Parallel (Balanced) Above Ground Plane For D << d, h_{1}, h_{2} Z_{0}= (276/ε^{½}) log_{10}{(2D/d)[1+(D/2h)^{2}]^{½}} 
Two Conductors in Parallel (Balanced) Different Heights Above Ground Plane For D << d, h_{1}, h_{2} Z_{0}= (276/ε^{½})log_{10}{(2D/d)[1+(D^{2}/4h_{1}h_{2})]^{½}} 
Single Conductor Between Parallel Ground Planes For d/h << 0.75 Z_{0}= (138/ε^{½}) log_{10}(4h/πd) 
Two Conductors in Parallel (Balanced) Between Parallel Ground Planes For d << D, h Z_{0}= (276/ε^{½}) log_{10}{[4h tanh(πD/2h)]/πd} 
Balanced Conductors Between Parallel Ground Planes For d << h Z_{0}= (276/ε^{½}) log_{10}(2h/πd) 
Two Conductors in Parallel (Balanced) of Unequal Diameters Z_{0}= (60/ε^{½}) cosh^{1} (N) N = ½[(4D^{2}/d_{1}d_{2})  (d_{1}/d_{2})  (d_{2}/d_{1})] 
Balanced 4Wire Array For d << D_{1}, D_{2} Z_{0}= (138/ε^{½}) log_{10}{(2D_{2}/d)[1+(D_{2}/D_{1})^{2}]^{½}} 
Two Conductors in Open Air Z_{0}= 276 log_{10}(2D/d) 
5Wire Array For d << D Z_{0}= (173/ε^{½}) log_{10}(D/0.933d) 
Single Conductor in Square Conductive Enclosure For d << D Z_{0}≈ [138 log_{10}(ρ) +6.482.34A0.48B0.12C]/ε^{½} A = (1+0.405ρ^{4})/(10.405ρ^{4}) B = (1+0.163ρ^{8})/(10.163ρ^{8}) C = (1+0.067ρ^{12})/(10.067ρ^{12}) ρ= D/d 
Air Coaxial Cable with Dielectric Supporting Wedge For d << D Z_{0}≈ [138 log_{10}(D/d)]/[1+(ε1)(θ/360)]^{½}) ε = wedge dielectric constant θ= wedge angle in degrees 
Two Conductors Inside Shield (sheath return) For d << D, h Z_{0}= (69/ε^{½}) log_{10}[(ν/2σ^{2})(1σ^{4})] ν = h/d σ = h/D 
Balanced Shielded Line For D>>d, h>>d Z_{0}= (276/ε^{½}) log_{10}{2ν[(1σ^{2})/(1+σ^{2})]} ν = h/d σ = h/D 

Two Conductors in Parallel (Unbalanced) Inside Rectangular Enclosure For d << D, h, w ∞ Z_{0}= (276/ε^{½}) {log_{10}[(4h tanh(πD/2h)/πd) ∑ log_{10}[(1+μ_{m}^{2})/(1ν_{m}^{2})]} ^{m=1} μ_{m}=sinh(πD/2h)/cosh(mπw/2h) ν_{m}=sinh(πD/2h)/sinh(mπw/2h) 